(Respiration) how water vapor affects barometric pressure

hello folks, let me apologize if this was addressed before, but every time I try to search for partial pressure of oxygen and water, I get problems unrelated to mine. I also searched Wikipedia and Google and found their explanations a bit too obtuse. But it's possible I'm just doing a bad job of research, if so, please feel free to point me to the right link... otherwise, I hope you can help me here.

When calculating the actual partial pressure of oxygen entering an alveoli, we take the following steps...

We first calculate barometric pressure (PB): usually 760 mmHg at sea level

We then calculate the vapor pressure of water in the lung (because the lung releases water in the respiratory space) at body temperature (PW): usually 47 mmHg at 37° C

Only at this point can we calculate the partial pressure of oxygen (PO2): which is 0.21 times the pressure of dry gas... 0.21 x 713 = 150.0 mmHg ← PO2

If you did not account for the water vapor reduction, you'd get a higher value of PO2 (160.0 mmHg) and you'd be wrong.

OK... I can do this much with my eyes closed.
my problem is why the water behaves as such, and some potential paradoxes that arise. I don't like to do math that seems magicky and I don't still get the actual physics behind it.

Dalton's Law of Partial Pressures: a simple equation that states PTOT = P1 + P2 + Pn so the total pressure is the sum of the partial pressures. Why is it that atmospheric pressure in the airway (or anywhere else) is reduced by water subtractively, but by oxygen fractionally? In other words, if PO2 in dry environment is 160 mmHg, why doesn't it remain this way when moistened and instead is reduced by its fraction? why aren't we just ADDING PW to the 760, or (if the pressure can't go up since the lung inspires and creates negative pressure) don't we just fractionate the water and all other gas pressures? is it the fact water vapor is "liquid"? (am I not correct in treating water vapor as a gas, and subject to Dalton's Law?)

The pressure of gases at 100°C: at this temperature, water's partial pressure would be 760 mmHg, equal to atmospheric pressure. Now, forgetting that human tissue boils at that temperature (imagine we were calculating this for a machine that is similar to a lung physically except it can handle the heat instead), wouldn't water's vapor pressure be such that there would be NO GAS PRESSURE? by the rules above, it seems at 100°C, the volume would be all water vapor and no oxygen, nitrogen, or argon pressure would be present... doesn't seem right. Keep in mind though temperature rise produces pressure rise, we're working in a constant pressure environment, so high temperature won't produce higher pressure if it can simply dissipate.

Staff: Mentor

The air in your lungs is roughly at force equilibrium with the air outside your body. So their total pressures have to be equal. But, in your lungs, water vapor has evaporated into the air, and comes to vapor-liquid equilibrium with the liquid water in your body at its equilibrium vapor pressure (at your body temperature). So in your lungs, the partial pressures of oxygen and nitrogen have to less than in the outside air in order to maintain the same total pressure. The way this happens is that, when you breathe in, you take in less dry air than is necessary to fill your lungs, and the water vapor makes up the difference. When the gases mix in your lungs (before you exhale), the partial pressures of oxygen and nitrogen become less than that of the dry air that you breathed in.

If somehow you increased your body temperature to 100C (ouch), and you could remain alive when this happens, then the liquid water in your body would be boiling and releasing water vapor at 1 atm into your lungs. There would be no oxygen or nitrogen, and you couldn't even breathe any in.

In short
1.why do we subtract vapor pressure from atmospheric pressure?
2.why is PO2 not added and instead multiplied/fractioned?
3.at 100°C, it seems PGASES would all equal ZERO since PW = PB

1. Why do we subtract water vapour pressure ....
The percentage of oxygen of dry air is stable and this works out to be 21%. Unless you know of an easier way to determine the partial pressure of oxygen without subtracting the water vapour pressure from the total atmospheric pressure, then this method does simplify the calculation.

2. Again, partial pressure of oxygen is 0.21 of the total pressure of dry air.

3.Depends doesn't it on whether one deals with an open or closed container.

If I get it, the atmosphere in the lung would be 1atm like it is outside the lung, so the water vapor pressure simply insinuates itself into the equation without increasing pressure, so whatever pressure it takes up, the rest of the gases will have a smaller share.

as for why the partial pressure of oxygen is 0.21, I only imagine this is because I was given the ratio... so I can use it to solve for gases. I imagine if i was given a quantity like 159.6 mmHg, I'd have to convert it to .21 first if I was to figure out the new value of oxygen...

Staff: Mentor

If I get it, the atmosphere in the lung would be 1atm like it is outside the lung, so the water vapor pressure simply insinuates itself into the equation without increasing pressure, so whatever pressure it takes up, the rest of the gases will have a smaller share.

as for why the partial pressure of oxygen is 0.21, I only imagine this is because I was given the ratio... so I can use it to solve for gases. I imagine if i was given a quantity like 159.6 mmHg, I'd have to convert it to .21 first if I was to figure out the new value of oxygen...

thank you so much for the insight guys, I really appreciate it!

As far as oxygen is concerned, coming into your lungs, it is going to have a mole fraction of ~0.21, but in the gas you exhale, the mole fraction is going to be <0.21, because of the oxygen that has been used by your body. There will also be more CO2 in what you exhale.